Lecture Note on Solid State Physics Bloch theorem and Energy bandMasatsugu Suzuki and Itsuko S. Suzuki Department of Physics, State University of New York at Binghamton, Binghamton, New York 13902-6000 (May 9, 2006) Abstract Here we consider a wavefunction of an electron in a periodic potential of metal. The translation symmetry of periodic potential is imposed on the wave function. The wave function of electrons is a product of a plane wave and a periodic function which has the same periodicity as a potential. These electrons are often called Bloch electrons to distinguish them from the ideally free electrons. A peculiar aspect of the energy spectrum of the Bloch electrons is the formation of energy band (allowed energy regions) and band gap (forbidden energy region). In this note we discuss the Bloch theorem using the concept of the translation operator, the parity operator, and the time-reversal operator in quantum mechanics. Our approach is similar to that used by S.L. Altmann (Band theory of metals: the elements, Pergamon Press, Oxford, 1970). This book is very useful in our understanding the concept of the Bloch theorem. The eigenvalue problems are solved, depending on the strength of the periodic potential (we use Mathematica 5.2 to solve the problems). The exact solution of the Kronig-Penny model is presented using Mathematica 5.2. We also discuss the persistent current of conducting metal ring in the presence of magnetic field located at the center (the same configuration as the Aharonov-Bohm effect) as an application of the Bloch theorem. Content 1. Translation operator 1.1 Analogy from the classical mechanics for x 1.2 Analogy from the classical mechanics for p 1.3 Infinitesimal translation operator in the position basis 1.4 Momentum operator p 1.5 The finite translation operator 2. Parity operator 2.1 Property 2.2 Commutation relation 2.3 Parity operator on electron-spin state 3. Time- reversal operator 3.1 Definition 3.2. Property 3.3 Time-reversal operator on electron-spin state 4. Bloch theorem 4.1 Derivation of the Bloch theorem 4.2 Symmetry of Ek and E-k: the time-reversal state 4.3 Kramers theorem for electron- spin state 14.4 Parity operator for symmetric potential 4.5 Brillouin zone in one dimensional system 4.6 Bloch wavefunction 4.7 Properties of energy band 5. Solution of the Schrdinger equation 5.1 Secular equation 5.2 Solution for the simple case 5.2.1 U G < 0 5.2.2 U G > 0 5.2.3 Probability of finding electrons ((Mathematica 5.2)) 5.3 Eigenvalue problem for the complicated case ((Mathematica 5.2)) 5.4 Energy dispersion curves 5.5 Bragg reflection at the boundary of the Brillouin zone 5.5.1 1D system 5.5.2 2D system 6. Kronig-Penny model as an application of the Bloch theorem 6.1 Secular equation 6.2 Energy dispersion relation ((Mathematica 5.2)) 7. Theory of persistent current in conducting metallic ring 7.1 Model similar to the Aharonov-Bohm effect 7.2 Derivation of energy eigenvalues as a function of magnetic flux 7.3 Energy eigenvalues and persistent current density as a function of magnetic flux ((Mathematica 5.2)) 8. Conclusion References Appendix1 1.1 Translation operator1 Analogy from the classical mechanics for x (a ) in quantum mechanics, Here we discuss the translation operator T ( a) , ' =T(1)or (2) In an analogy from the classical mechanics, it is predicted that the average value of x in the old state plus the x-displacement a in the new state ' is equal to that of x under the translation of the system ' = x + a , 'x or + ( a) x ( a) = x T + a , T or . + (a ) x (a ) = x T + a1 T (3) Normalization condition: + ( a) . ' = 'T2 + (a)T (a ) = , ' ' = T or + (a )T (a) = 1 . T (a) is an unitary operator]. [T From Eqs.(3) and (4), we have (a) = T (a )( x (a) x (a) , T + a) = T + aT x or the commutation relation: (a )] = aT (a) . ,T [x From this, we have (a ) x = T ( a) x (a) x = ( x + a)T (a ) x . T x + aT x (a) x is the eigenket of x with the eigenvalue (x+a). Thus, T(4)(5)or ( a) x = x + a , T or + (a)T (a ) x = T + ( a) x + a = x . T When x is replaced by x-a in Eq.(7), we get + (a ) x , xa =T or ( a) . xa = xT Note that (a ) = x a = ( x a ) . x' = xT1.2(6) (7) (8) (9) (10)Analogy from the classical mechanics for p in the new state ' is equal to the average value of p in The average value of p' = p , ' pthe old state under the translation of the system (11) or + (a ) p ( a) = p T , T or + (a) p (a ) = p T . T So we have the commutation relation . (a), p ] = 0 [T From this commutation relation, we have (a ) p = T (a ) p ( a) p . T p = pT p (a) p is the eigenket of p associated with the eigenvalue p. Thus, T(12)1.3Infinitesimal translation operator3We now define the infinitesimal translation operator by i G dx , (dx) = 1 T(13) is called a generator of translation. The dimension of G is that of the linear G momentum. (dx) satisfies the relations: The operator T , + (dx )T (dx) = 1 T (14) + (dx ) x (dx ) = x T + dx , T or (dx ) T (dx) x (dx ) , T = dxT x(15) (16)and, (dx ), p ] = 0 [T Using the relation (14), we get i G dx )+ (1 i G dx) = 1 , (1or + i G + dx)(1 i G dx ) = 1 + i (G + G )dx + O[(dx) 2 ] = 1 , (1or + = G . G is a Hermite operator. Using the relation (15), we get The operator G i i i G dx) = dx1 + O (dx) 2 , (1 G = dx (1 x dx ) (1 G dx ) x ori ]dx = dx1 , , G [x(17)or ] = i 1 . [ x, G Using the relation (16), we get i G dx, p . ] = 0 [1(18)Then we have , p . ] = 0 [G From these two commutation relations, we conclude that =p , G and i p (dx) = 1 dx . T(19)(20) and the momentum operator p obeys the We see that the position operator x commutation relation4] = i 1 , [ x, p which leads to the Heisenbergs principle of uncertainty.(21)1.4 in the position basis Momentum operator pT (x ) = T (x) dx ' x' x' = dx' x '+x x ' = dx ' x' x'x = dx' x' ( x'x) . We apply the Taylor expansion: ( x'x ) = ( x' ) xSubstitution: ( x' ) . x ' ( x ' )] = x' i p x) . = dx ' x' [ x' x x' ] = x dx' x ' x ' = (1 x' x ' Thus we have = p dx' x ' x' , i x ' = xp dx ' x x ' x' = dx' ( x x' ) x' = x . x' x' i i i x We obtain a very important formula x p = x . (22) i x * = dx x = dx x p xp x = dx x x . i i x i i xT (x ) = dx ' x' ( x'x ) = dx ' x' [ ( x ' ) x The finite translation operator (a) corresponding to a finite translation a? We find it by the What is the operator T , a/N following procedure. We divide the interval a into N parts of size dx = a/N. As N becomes infinitesimal. i a T (dx ) = 1 p ( ) . N Since a translation by a equals N translations by a/N, we have a i i p (a) = Lim[1 ( )]N = exp( p a) . T N N ! !1.5N0aa ( ) N5Fig.1 The separation of a divided by N, which becomes infinitesimally small when N . Here we use the formula 1 1 lim (1 + ) N = e , lim (1 ) N = e 1 , N N N N N ax ax lim [(1 ) ax ]ax = lim (1 ) N = (e 1 )ax = e ax . N N N N In summary, we have (a) = exp( i p a) . T It is interesting to calculatei a p i a p(23) + ( a) x ( a) = e x T e , T by using the Baker-Hausdorff theorem: x x2 x3 exp( Ax) B exp( Ax ) = B + [ A, B] + [ A,[ A, B]] + [ A,[ A,[ A, B]]] + ... 1! 2! 3! When x = 1, we have )B ) = B , B ,[ A , B ,[ A ,[ A , B exp( A + 1[A ] + 1 [A ]] + 1 [ A ]]] + ... exp( A 1! 2! 3! Then we have i i a a p p i i i + . T ( a) = e x e +[ p a, x ] = x + a[ p , x ] = x + a =x + a1 =x T (a ) x i So we confirmed that the relation . + (a ) x ( a) = x T + a1 T holds for any finite translation operator.2 2.1 Parity operator1 Property zRH (right-handed) new y y LH (left-handed) x new z new xFig.2 Right-handed (RH) and left-handed (LH) systems.6 : parity operator (unitary operator) , ' = or+ . ' = (24) (25) in the new state ' is opposite to to that in the old We assume that the average of xstate ' = x , ' xor+x = x , or +x = x . The position vector is called a polar vector. We define the normalization by , + = =1 ' ' = or . + =1 Thus the parity operator is an unitary operator. From Eqs.(26) and (27), , + x =0 x or x = x x = x x . x x is the eigenket of x with the eigenvalue (x). Thus or x = x , or x = x = x , or . 2 =1 and , + =1 2 =1 Since + = , or + = . So the parity operator is a Hermite operator.(26)(27)(28)(29)(30) p = dx' x' x' p = dx' x' x' p = dx' x x ' p 1 ipx ' = dx x ' exp( ) = dx x 2 Note that x' = -x and dx' = dx. Then we have 1 ipx exp( ) = dx x x p . 2 7 p = p , (31)and p =p p . p So we have p p = p p = pp , or p =p p = p p . pThus we have , p +p =0 or + p = p . Thus the linear momentum is called a polar vector.2.2(32)Commutation relation Here we show the commutation relation between the parity operator and several (a) . The orbital angular momentum L (the x axis component) is operators including T x z or =x ]=0 ) holds p y y px . The commutation relation ( [ , L +L =L defined by L zz z z x , and p y are odd under parity. Similar commutation relations hold , y , p valid, since x and general angular momentum J and : +S =S for the spin angular momentum S ( a) ; . We show that there is a commutation relation between +J =J and T x. (a )] = 0 ( a) = T + (a) T ,T , or [ x x x i i + i (a) ( a) = T + (a ) , +T = + exp( p a ) = exp( p a ) = exp( p a) = T x x xor (a ) = T + (a) +T , x xor ( a) = T + (a) T , x x + = . since (33) = 1 p 2 + V (x ) . Here we assume that the potential is The Hamiltonian is given by H 2m ) = V ( x ) . Then we have the commutation relation symmetric with respect to x = 0: V ( x + + , since ]=0 or [ , H V (x ) = V ( x ) = V ( x ) and H =H + p 2 = ( p )2 = p 2. In conclusion, we have the following commutation relations. , , and . +S =S +L =L +J =J (1) + ( a) = T (a) T . (2) x x for H =p +H =H 2 /(2m) + V ( x ) , only if V ( x ) = V ( x ) . (3) 82.3Parity operator on electron-spin state Electrons has a spin (s = 1/2). The spin angular momentum is S (= / 2 ) and the spin magnetic moment is given by s = (2B/ )S, where B (= e /2mc) is a Bohr magneton. We now consider how the electron-spin state changes under the parity operator. The spin of electron commutates with . Since =S : S 1 = S S , operator S z z + = + = S + , where + is the spin-up state. So the state + is the S z z 2 with the eigenvalue / 2 , or + = + . Similary we have eigenket of = = S , where is the spin-down state. So the state is the S z z 2 with the eigenvalue / 2 , or = . eigenket of In conclusion the spin state remains unchanged under the parity operator: + = + and = . (34)Time-reversal operator1 Definition The time reversal is an odd kind of symmetry. It suggests that a motion picture of a physical event could be run without the viewer being able to tell something is wrong. We now consider the Schrdinger equation i (t ) = H (t ) . t Suppose that (t ) is a solution. We can easily verify that (t ) is not a solution because of the first-order time derivative. However, i * (t ) = H * * (t ) = H * (t ) . t When t t , we have i * (t ) = H * (t ) . t This means that * (t ) is a solution of the Schrdinger equation. The time reversal state is defined by (t ) = * (t ) . (35) 3. 3.1If we consider a stationary state, (t ) = ei Et (0) , * ( t ) = eor Eti * ( 0) , (t ) = [e ori Eti Et (0)] = ei Et * (0) ,e (0) = e * (0) , where (0) = * (0) = K (0) and K is an operator which takes the complex conjugate.i Et93.2. Property ~ ) are The state before the time reversal ( ) and the state after the time reversal ( related through the relation ~ = , (C + C ) = C *1 + C * ) . The time-reversal operator acts only to where 1 2 2the right because it entails taking the complex conjugate. The inner product of the time~ ~ = and is defined by reversal states = ~ ~ * (36) = = . One can then show that the expectation operators must satisfy the identity ~ 1 ~ + = A *. A = A , then we have A 1 = A Suppose that ~ 1 ~ ~ ~ ~ ~ ~ * ~A = . A A = A = A = If = , we have ~A ~ . = (38) A In conclusion, most operators of interest are either even or odd under the time (+: even, -: odd). A 1 = A reversal. (i is a pure imaginary, 1 is the identity operator). i 1 = i1 (1) p = p ). p 1 = p : ( (2) p 1 = p 2 2 . (3) (4) (5) (6) (7) (37) r 1 r = r ). : ( = r V (r 1 = V (r ) is a potential). ) ) : ( V (r (S is the spin angular momentum). S 1 = S (8)2 H 1 = H , when H = p ) and V ( x ) is a potential energy. The relation + V (x 2m ) . is independent of the form of V ( x T ( a) = T ( a) . T (a ) 1 = T (a) or x x x x3.3Time reversal operator on electron-spin state We now consider how the electron-spin state change under the time-reversal operator. and = S , we have S 1 = S S Since z z z z + = + . + = S S z z 2 with an eigenvalue / 2 . Then we The time reverse state + is the eigenket of S z + = , where is a phase factor (a complex number of modulus unity). Here have can be expressed by we choose = 1. In this case, 10 = i , yK (39) is an operator which takes the complex conjugate and y is a Pauli spin where K y + = i and y = i + . First we calculate operator. Note that (C + + C ) = i (C + + C ) = i K (C * + + C * ) + y + y + y + + C y ) = C+ C + . = i (C + * * * * again to the above where C1 and C2 are arbitrary complex numbers. We try to apply state (C * C * + ) 2 (C + + C ) = (C * C * + ) = i yK + + + y (C+ C + ) = i[(C+ y C y + )] = i = i[(C+ (i) + Ci + )] = (C+ + + C ) or . 2 = 1 .(40)4Bloch theorem Felix Bloch entered the Federal Institute of Technology (Eidgenssische Technische Hochschule) in Zrich. After one year's study of engineering he decided instead to study physics, and changed therefore over to the Division of Mathematics and Physics at the same institution. After Schrdinger left Zrich in the fall of 1927 he continued his studies with Heisenberg at the University of Leipzig, where he received his degree of Doctor of Philosophy in the summer of 1928 with a dissertation dealing with the quantum mechanics of electrons in crystals and developing the theory of metallic conduction.By straight Fourier analysis I found to my delight that the wave differed from the plane wave of free electrons only by a periodic modulation. This was so simple that I did not think it could be much of a discovery, but when I showed it to Heisenberg, he said right away; Thats it!! (F. Bloch, July, 1928) (from the book edited by Hoddeson et al.2). His paper was published in 1928 [F. Bloch, Zeitschrift fr Physik 52, 555 (1928)]. There are many standard textbooks3-10 which discuss the properties of the Bloch electrons in a periodic potential.Derivation of the Bloch theorem We consider the motion of an electron in a periodic potential (the lattice constant a). The system is one-dimensional and consists of N unit cells (the size L = Na, N: integer). ) = V ( x + a1 ) , V (x + , ( )x ( )=x T + 1 T (41)x4.1x ( ) x = x+ T x,(42) (43) i p (x) = 1 xx , T x11where l is any finite translation (one dimensional) and x is the infinitesimal translation. a is the lattice constant. The commutation relations hold , (x), p x] = 0 [T x and . (x), p x2 ] = 0 [T x Therefore the kinetic energy part of the Hamiltonian is invariant under the translation. When = a (a is a period of potential V(x)), , + ( a) x ( a) = x T + a1 T x x ) = V ( x + (a)V ( x ( a) = V ( x )T + a1 ) . TxxThus we have , ,T (a )] = 0 [H x or + (a ) H T (a ) = H . T x x The Hamiltonian is invariant under the translation with a. (a) x = x + a and T + (a ) x = x a or T + ( a) x = x + a , Since T we have Tx + ( a) = Tx (a ). + (a) is not a Hermite operator. So T (44)(45) and T (a) for the system with a We consider the simultaneous eigenket of H x . ,T (a )] = 0 periodicity of L = Na (there are N unit cells), since [ Hx =E , H k k k and ( a) = 1 , T x k k p or (46) (47)N (a)] [T x kN1 = p k .(48)Note that ( a) x = x + a , T x (a)]N x = x + Na = x (periodic condition). [T x Thus we have p N = 1, or 2s 2as p = exp(i ) = exp(i ) = exp(ika ) , N Na with 2 2 k= s= s (s: integer). Na L(49)(50)12Therefore, we have (a) = e ika . T x k k (a ) with the eigenvalue e ika . The state k is the eigenket of T x or (a) = exp(ika) x , xT x k k (a ) = x a , xTx(51)x a k = e ika x k ,(52) (53) (54)or k ( x a ) = e ika k ( x) . By changing for a to a, we have k ( x + a) = eika k ( x ) . This is called as the Bloch theorem.4.2Symmetry of Ek and E-k: the time-reversal state is invariant under the time-reversal operator (this We assume that the Hamiltonian H H =H . Then the state k is the simultaneous assumption is valid in general): and : eigenket (the Bloch state) of H = E and T (a) = e ika . H (55)k k k x k k~ = = H = E , the time-reversal state is also the Since H k k k k k k with the energy eigenvalue Ek. Since [T , (a ), ]=0 eigenket of H x ~ =T ~ . (a ) ( a ) = T (a ) = (e ika ) = eika = eika T x k x k x k k k k ~ The time-reversal state k = k is the eigenket of T (a) with the eigenvalue eika. Sox~ is different from the state and coincide with the state the state k k k , where ~ =E ~ . H k k k In conclusion, the property of Ek = E-k is a consequence of the symmetry under the time reversal: = . (56) (1) k k = (2) Both states ( and ) are degenerate states with the same energykkkeigenvalue: Ek = E k .(57)4.3Kramers theorem for electron-spin state We consider how the electron-spin state changes under the time reversal. The is invariant under time reversal, [ H . Let , ]=0 Hamiltonian H and k ,s k , s = k , s be the simultaneous eigenket of H and S z ( [ H , S z ] = 0 ) and its time-13 reversed states, respectively. H k , s = Ek , s k , s , where Ek,s is the eigenket with the wavenumber k and spin state s (s = up or down). H k , s = H k , s = Ek , s k , s = Ek , s k , s = Ek , s k , s . with the eigenvalue Ek,s. On the other hand, It follows that is the eigenket of H k,s k , s = k , s is the eigenket of H with the eigenvalue E-k,-s. Therefore Ek,s is equal (half-integer), 2 = 1 and k , s are orthogonal. This means to E-k,-s. When k,s that and k , s (having the same energy Ek,s) must correspond to distinct states k,s (degenerate) [Kramers theorem]. In order to prove this orthogonality, we use the formula ~ ~ * = = , where ~ 2 = k , s , = = ( k ,s ) = k ,s = k ,s , = k ,s = . ~ 2 2 Since k , s = k , s , we have = k , s = k , s . Then or ~ ~ = = , ~ k ,s = 0 ,indicating that for such systems, time-reversed states are orthogonal. In conclusion, when the effect of spin on the energy eigenket is taken into account (1) (58) k , s = k , s . (2) Both states ( and k , s ) are degenerate states (the same energy k , s = k , s but different states): Ek , s = E k , s , or Ek , = E k , and Ek , = E k , . (59)4.4Parity operator for symmetric potential What is the effect of the parity operator on the eigenket k ? Using the following relations (a) = e ika , T x k k + T (a) = T (a ) , x xwe have (a ) = T + (a ) T k = e ika k , x k x or + (a ) k = e ika k , T x or 14 (a ) k = e ika k . T x When a is changed to a in the above equation, we get (a) k = eika k . T (60) x (a ) with the eigenvalue eika . k is the eigenket of T In other words, the state x Here we consider the limited case that the potential energy V(x) is an even function of commutes with . In other , ) = V ( x ) . Then the Hamiltonian H : [H ] = 0 x:or V ( x is invariant under the parity operation. The state words, H is a simultaneousk,s ika and T (a ) (a ) : H and T k ,s . eigenket of H k , s = Ek , s k , s x k ,s = e x k,s = H H k , s = Ek , s k , s . with Ek,s and T (a) with eika. The state k , s is the simultaneous eigenket of H Thus x k , s coincides with k , s with E-k,s. Therefore we can conclude that Ek,s = E-k,s; Ek, = E-k, and Ek, = E-k, . Brillouin zone in one dimensional system We know that the reciprocal lattice G is defined by 2 G= n , (n: integer). a When k is replaced by k + G, k + G ( x + a ) = ei ( k + G ) a k + G ( x) = eika k + G ( x) ,4.5(61)since eiGa = ei 2n = 1 . This implies that k + G ( x) is the same as k ( x) . k +G ( x) = k ( x) . (62) or the energy eigenvalue of k + G ( x) is the same as that of k ( x) , Ek + G = Ek . (63) Note that the restriction for the value of s arises from the fact that k + G ( x ) = k ( x ) . 2s 2s 2 s = = ( ), k= L Na a N where N N s . 2 2 . There are N states in the first Brillouin zone. a When the spin of electron is taken into account, there are 2N states in the first Brilloiun zone. Suppose that the number of electrons per unit cell is nc (= 1, 2, 3, ). Then the number of the total electrons is ncN. (a) nc = 1. So there are N electrons. N/2N = 1/2 (band-1: half-filled). (b) nc = 2. 2N/2N = 1 (band-1: filled). (c) nc = 3. 3N/2N = 1.5 (band-1: filled, band-2: half-filled). (d) nc = 4. 4N/2N = 2 (band-1: filled, band-2: filled). The first Brillouin zone is defined as k 15When there are even electrons per unit cell, bands are filled. Then the system is an insulator. When there are odd electrons per unit cell, bands are not filled. Then the system is a conductor.4.6Bloch wavefunction Here we assume that k ( x) = eikxuk ( x) ,(64) k ( x a ) = eikx e ikauk ( x a) = e ika eikxuk ( x a) , which should be equal to e ika k ( x) = e ika eikxuk ( x) , or uk ( x a) = uk ( x ), (65) which is a periodic function of x with a period a. The solution of the Schrodinger equation for a periodic potential must be of a special form such that k ( x) = eikxuk ( x) , where uk ( x + a) = uk ( x) . In other words, the wave function is a product of a plane wave and a periodic function which has the same periodicity as a potential Here we consider the 3D case. The solutions of the Schrdinger equation for a periodic potential must be of a special form: k (r ) = uk (r )eik r (Bloch function), (66) where u k (r ) = u k (r + T ) . (67) Bloch functions can be assembled into localized wave packets to represent electrons that propagate freely through the potential of the ion cores. T is any translation vectors which is expressed by T = n1a1+n2a2+n3a3 (n1, n2, n3 are integers, a1, a2, a3 are fundamental lattice vectors). From Eq.(67), uk (r ) can be expanded as follows. (Fourier transform) uk (r ) = Ck G e iG r . (68)Gwhere G is the reciprocal lattice vector. We use the same discussion for the periodic charge density in the x-ray scattering. Then the wave function in a periodic potential is given by k (r ) = Ck G e i ( k G )r = Ck e ik r + Ck G ei ( k G )r + ...Gor k (r ) = ... + Ck 2G ei (k 2G )r + Ck G e i ( k G )r + Ck eik r + Ck + G ei (k + G )r + Ck + 2G ei (k + 2G )r + .... (69) The eigenvalue-problem =E , H or H k ( x ) = Ek k ( x ) . k k k Ek is the eigenvalue of the Hamiltonian and has the following properties. (i) Ek = Ek + G . (ii) Ek = Ek . (70) The first property means that any reciprocal lattice point can serve as the origin of Ek. The relation Ek = Ek is always valid, whether or not the system is centro-symmetric.16The proof of this is already given using the time-reversal operator. The proof can be also made analytically as follows. H k ( x ) = Ek k ( x ) , * * is Hermitian), H ( x ) = E ( x ) ( Hk k korH k ( x) = E k k ( x ) . From the Bloch theorem given by k ( x a ) = e ika k ( x ) , or k ( x) = eikxuk ( x) , and k * ( x) = e ikx uk * ( x ) , we have k * ( x a ) = e ik ( x a )uk * ( x a ) = e ik ( x a )uk * ( x ) = e ika k * ( x) , or k * ( x a) = e ika k * ( x) .* * * (a) with Thus the wave functions k ( x) and k ( x) are the same eigenfunctions of T xthe same eigenvalue e ika . Thus we have k * ( x ) = k ( x) , with Ek = Ek . What does this relation mean? k (r ) = Ck G ei (k G )r = Ck eik r + Ck G ei ( k G )r + ...(71) k (r ) =*GCk G e i (k G )r ,*Gor k (r ) =* GC k G e* i ( k + G ) r=GC k + G ei (k G )r .*Then we have the relation * C k + G = Ck G , or * Ck G = C k + G . 4.7 (i)(72)Properties of energy band Ek = Ek + G We consider the case of an infinitely small periodic potential. The curve Ek is practically the same as in the case of free electron, but starting at every point in reciprocal lattice at G = (2/a)n (n: integer). We have Ek+G = Ek, but for the dispersion curves that have a different origin. ((Mathematica 5.2))17f2 2 n ; rule1 m 1, 1, a 1 ; g f . rule1; 2m a s1 Plot Evaluate Table g, n, 3, 3 , k, 3 , 3 , PlotRange 3 , 3 , 0, 125 , Prolog AbsoluteThickness 2.0 , PlotStyle Hue 0 , Hue 0 , Background GrayLevel 0.8 2k120 100 80 60 40 20-7.5 -5-2.52.557.5Graphics Fig.3 The energy dispersion (Ek vs k) of electrons in the weak limit of periodic potential (the periodic zone scheme), where Ek = Ek + G . m 1. a 1. 1. G 2n (n = 0, 1, ). (ii) E k = EkFig.4 The relation of E-k = Ek in the reciprocal lattice plane. k = /a is the boundary of the first Brillouin zone (|k| /a).It follows that from the condition ( Ek = E k ), in Fig.4, E(1) = E(2). On taking 0, the group velocity defined by [ E (2) E (1)] / 2 reduces to zero (dk/dk 0). On applying the periodicity condition Ek = Ek + G this result can immediately be extended as follows. dk/dk 0 at k = 0, 2/a, 4/a,., We now consider the value of this derivative at the Brillouin zone boundary. From the condition Ek = E k , E(3) = E(4). From the condition Ek = Ek + G , E(3) = E(5). Therefore, we have E(4) = E(5). On taking 0, the group velocity at the boundary of Brillouin zone is defined as [ E (5) E (4)] / 2 , which reduces to zero (dk/dk 0). 18In conclusion, the group velocity (dEk/dk) is equal to zero at k = 0, G/2, G, 3G/2, 2G,. (see the books written by S.L. Altmann3 for more detail.) 5 5.1 Solution of the Schrdinger equation Secular equation We consider the Schrdinger equation of an electron in a periodic potential U(x) with a period a. 2 d2 [ + U ( x )] k ( x) = E k ( x) , (73) 2m dx 2 where U ( x ) = U G e iGx [(G = n (2/a), n: integer)], (74)GwithU G = U G ,* k ( x) =withCk G ei ( k G ) x = Ck e ikx + Ck G ei ( k G ) x + ... ,G * C k G = C k +G , 22m Here we note that GCk G (k G ) 2 ei ( k G ) x + G' U G 'e iG ' x G Ck G e i ( k G ) x = EGCk G e i ( k G ) x .I =G'U G 'e iG ' xCk G eGi (k G) x=G G'U G 'Ck G e iG ' x ei ( k G ) x .For simplicity, we put G" = G + G ' or G ' = G"G I= U G " G C k G e i ( k G ") x = U G G 'Ck G 'ei ( k G ) x ,G G" G' Gwhere we have a replacement of variables: G" G, G G ' in the second term. Then the Schrdinger equation is 2 G2m"Ck G (k G ) 2 ei ( k G ) x +G G'U G G 'Ck G 'e i ( k G ) x = EGCk G e i ( k G ) x ,or22m When k k + G$[(k G ) 2 E ]Ck G + ! U G G 'C k G ' = 0 .G'2(2mk 2 E )Ck + # U G G 'Ck + G G ' = 0 .G'%Here we put k =22m [k G E ]Ck G + & U G G 'Ck G ' = 0 ,G'k2 .or19[k G E ]Ck G + (... + U 4G Ck 5G + U 3G Ck 4G + U 2G Ck 3G + U G Ck 2G + U 0Ck G + + U G Ck + U 2G Ck + G + U 3G Ck + 2G + U 4G Ck + 3G + ...) = 0 When k k + G in Eq.(75) [k E ]Ck + (... + U 4G Ck 4G + U 3G Ck 3G + U 2G Ck 2G + U G Ck G + U 0Ck + + U G Ck + G + U 2GCk + 2G + U 3GCk + 3G + U 4GCk + 4G + ...) = 0 When k k + 2G in Eq.(75) [k + G E ]Ck + G + (... + U 4G Ck 3G + U 3GCk 2G + U 2G Ck G + U GCk + U 0Ck + G + + U G Ck + 2G + U 2G Ck + 3G + U 3G Ck + 4G + U 4G Ck + 5G + ...) = 0 (77) The secular equation is expressed by U G U 2G U 3G U 4G U 5G U 6G C k + 3G k + 3G E UG U G U 2G U 3G U 4G U 5G Ck + 2 G k + 2G E U 2G UG U 2G U 3G U 4G Ck + G k + G E U G U 3G U 2G UG U G U 2G U 3G Ck = 0 , k E U 4G U 3G U 2G UG U G U 2G Ck G k G E U 5G U 4G U 3G U 2G UG U G C k 2G k 2G E U 6G U 5G U 4G U 3G U 2G UG k 3G E Ck 3G with U0 = 0 for convenience, where we assume that Ck+mG =0 for m = 4, 5, 6,. (75), (76).5.2Solution for the simple case Now we consider the simplest case: mixing of only the two states: k and k G (k /a. k-G = -/a, G2/a). Only the coefficients Ck and Ck-G are dominant. Ck 0 UG k E = . 0 UG k G E Ck G From the condition that the determinant is equal to 0, 2 ( k E )( k G E ) U G = 0 , or* (78)E=k + k G (k k G )2 + 4 U G222.(79)Now we consider that k k G ( k = k G with k (k E ) 2 U G = 0 , or/a, Bragg reflection)20E = k U G .Note that the potential energy U(x) is described by U ( x ) = U 0 + U G e iGx + U G e iGx = U 0 + 2U G cos(Gx ) , where we assume that UG is real: * U G = U G = U G . At k = G = 2/a only the coefficients Ck-2G and Ck are dominant. In this case we have the secular equation only for Ck-2G and Ck.. U 2G Ck k E 0 . = U 2G k 2G E Ck 2G 0 The condition of det(M) = 0 leads to k EU 2GU 2G = 0. k 2G E2*Since k = k 2G , we have (k E ) 2 U 2G = 0 , orE = k U 2G .5.2.1UG < 0 For E = k + U G = k U G (upper energy level)UG UG UG UG Ck Ck G= 0 0 ,or Ck = 1 . Ck G Then the wave function is described by k ( x) = Ck eikx + Ck G ei ( k G ) x = Ck [eikx ei ( k G ) x ] = 2iCk eorGx ) (upper energy level). 2 For E = k U G = k + U G (lower energy level)i(k G )x 2sin(Gx ), 2 k ( x) = 4 Ck sin 2 (2 2 UG UG UG UG Ck Ck G= 0 0 ,or Ck =1. Ck G The wave function is described by21 k ( x) = Ck eikx + Ck G ei ( k G ) x = Ck [eikx + ei ( k G ) x ] = 2Ck eori(k G )x 2cos(Gx ), 2 k ( x) = 4 Ck cos2 (2 2Gx ) (lower energy level). 25.2.2UG > 0 k ( x) = 4 Ck cos2 (2 2Gx ) 2for E = k + U G (upper energy level), for E = k U G (lower energy level).and k ( x) = 4 Ck sin 2 (2 2Gx ) 25.2.3 Probability of finding electrons ((Mathematica 5.2)) Comparison of the two standing wave solutions at k /a is presented. Note that the wave motion is in phase with the lattice.f1 Cos x 2; f2 Sin x 2; f3 1.2 Cos 2 x ; Plot f1, f2, f3 , x, 2, 2 , PlotStyle Table Hue 0.4 i , i, 0, 3 , Prolog AbsoluteThickness 3 , Background GrayLevel 0.81 0.5 -2-1 -0.5 -1 -1.5 -212Graphics Fig.5 At k = /a, Bragg reflection of the electron arises, leading to two possible charge distributions f1(x) and f2(x). The case of UG1;a[x_]:=a[x+2]/;x